# Implicit function theorem-critical points

Let $$f:U\rightarrow \mathbb{R}$$, where $$f$$ is $$C^{1}$$ and $$U\subset \mathbb{R}^{n}$$ is a open set. If $$f$$ has no critical points, show that $$f(A)$$ is a open set in $$\mathbb{R}$$, for all open set $$A\subset U$$.

My ideia is: Let $$A\subset U$$ arbitrary open set. For each point $$z\in A$$, we have $$\frac{\partial f}{\partial x_{i}}(z)\neq 0$$, for some $$i\in \{1,...,n\},$$ and usin the Implicit function theorem, $$f$$ converts a straight line segment parallel to the i-th axis, containing $$z$$ and small enough to be contained in $$A$$, injecting and monotonically over a range containing $$f(z)$$ and contained in $$f(A)$$, then $$f(A)$$ is a open set. Since $$A\subset U$$ is a arbitray open set, then this holds true for any open subset cointained in $$U$$.

## 1 Answer

The implicit function theorem is a particular case of the constant rank theorem, which says that if you have a function $$f$$ whose differential has constant rank (in your case, $$rank\, Df = 1$$) near a point $$p$$ in the domain, then there are open neighbourhoods $$U$$ of $$p$$ and $$V$$ of $$f(p)$$ and diffeomorphisms $$u:T_p M \to U$$, $$v:T_{f(p)}\to V$$ such that $$f(U)\subseteq V$$.

For each value $$f(p)$$ in $$f(A)$$ there is, at least, a point $$p$$ in $$A$$ where the requirements of the theorem hold. This yields an open neighbourhood $$V_{f(p)}$$ for each value $$f(p)$$ in $$A$$.

Thus, every value $$f(p)$$ in $$f(A)$$ has an open neighbourhood $$V_{f(p)}$$ that is contained in $$f(A)$$. So, since $$f(A)$$ is the union $$\cup_{p\in A} V_{f(p)}$$ of these open sets, $$f(A)$$ is itself open, as we wanted to show.